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svlad2 [7]
2 years ago
10

I need help please with this chemistry work ​

Chemistry
1 answer:
Marysya12 [62]2 years ago
8 0

Answer:

question 1: 3

question 2: the number of Valence electrons in the atom

hope it helps

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If 8.6 g of ch4 and 5.9 g of o2 react, what is the mass, in grams, of h2o that is produced?
Alenkasestr [34]
This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water. 

REMEMBER, you have to write and balance the chemical equation before you can do any of that work. 
That happens to be CH4 + 2O2 => CO2 + 2H2O
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3 years ago
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Answer: Objects with like charge repel each other.

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2 years ago
Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

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8 0
2 years ago
How many milliliters of 5.0 M NaOH are needed to exactly neutralize 40. milliliters of 2.0 M HCl?
zubka84 [21]

Answer:

16mL

Explanation:

Using the following formula;

CaVa = CbVb

Where;

Where

Ca = concentration/molarity of acid (M)

Va = volume of acid (mL)

Cb = concentration/molarity of base (M)

Vb = volume of base (mL)

According to the information provided in this question;

Ca (HCl) = 2M

Cb (NaOH) = 5M

Va (HCl) = 40mL

Vb (NaOH) = ?

Using CaVa = CbVb

Vb = CaVa/Cb

Vb = 2 × 40/5

Vb = 80/5

Vb = 16mL

5 0
2 years ago
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